Question

How do you prove that the limit ${{x}^{2}}=0$ as x approaches 0 using the epsilon delta proof?

Answer 1

We need to prove that the limit ${{x}^{2}}=0$ as x approaches 0 using the epsilon delta proof. We start to solve the given question by defining the epsilon delta proof. Then, for the limit of ${{x}^{2}}=0$ , we need to show that for any $\varepsilon >0$ there is $\delta >0$ such that $\left| x-0 \right|<\delta$ and so, $\left| {{x}^{2}}-0 \right|<\varepsilon$.

Complete step-by-step solution:
We are given a limit and asked to prove that the value of the limit is zero as x approaches 0. We will be solving the given question using the epsilon delta proof.
The epsilon delta proof is given as follows,
Let us assume that the function f has a limit $L$ as the value of x approaches c.
The above lines can be written in the form of an equation as follows,
$\Rightarrow \displaystyle \lim_{x \to c}f\left( x \right)=L$
According to the epsilon delta, the above equation is valid only if for every any $\varepsilon >0$ there is $\delta >0$ such that $\left| x-c \right| < \delta$ and so, $\left| f\left( x \right)-L \right| < \varepsilon$
Here,
$\varepsilon ,\delta$ are any positive numbers.
As per the given question, we need to prove that $\displaystyle \lim_{x \to 0}\text{ }{{x}^{2}}=0$
Let the value of $\varepsilon$ be greater than zero implies $\varepsilon >0$ and the value of $\delta$ be the minimum of $\varepsilon$ and one implies \delta =\min \left\\{ \varepsilon ,1 \right\\}
Let us consider,
$\Rightarrow \left| x-0 \right|=\left| x \right|<\delta$
From the above, we know that \delta =\min \left\\{ \varepsilon ,1 \right\\} implies $\delta \le 1$
Substituting the same in the above equation, we get,
$\Rightarrow \left| x-0 \right|=\left| x \right|<1$
From the above equation, we can conclude that
$\Rightarrow \left| {{x}^{2}} \right|=\left| {{x}^{2}} \right|<\left| x \right|.1=\left| x \right|$
From the values of the variables $\varepsilon ,\delta$ , we know that $\delta \le \varepsilon$
Following the same, we get,
$\Rightarrow \left| x \right|<\varepsilon$
From the above,
$\Rightarrow \left| {{x}^{2}}-0 \right|=\left| {{x}^{2}} \right|=\left| {{x}^{2}} \right| < \left| x \right| < \varepsilon$
We have found that for a value of $\varepsilon >0$ , there is a value of $\delta >0$ such that $\left| x-0 \right| < \delta$ that implies $\left| {{x}^{2}}-0 \right| < \varepsilon$
From the above,
$\therefore$ The value of $\displaystyle \lim_{x \to 0}\text{ }{{x}^{2}}=0$

Note: The value of the limit is the value that a particular function approaches as the index approaches some value. They help us to determine the tendency of the function at a particular point even though a function is not defined at the point.